1- The aim of compacting a fill is to fit as many soil solids into a given volum
ID: 1732585 • Letter: 1
Question
1- The aim of compacting a fill is to fit as many soil solids into a given volume as possible. Two soils are available as potential fill, one is classified as SW and the other SP. Which would make a better fill? Briefly explain why.
2- Soil A is classified as GW and soil B as CH. Which soil, A or B, has the highest: a) plasticity, ii) permeability.
3- A soil sample has a void ratio of 0.6 and a degree of saturation of 75%. Determine: a) The total unit weight of the soil sample. b) The shrinkage limit of the soil.
4- A trench 80 m long by 0.45 m wide by 1m deep is to be excavated in a clay soil at a site. An undisturbed cylindrical sample of the clay, with a diameter of 35 mm and length of 70 mm, is taken prior to digging the trench. The original mass of the sample is 101.5 g where mass of the sample after oven drying is 67.3 g. When the trench is dug the excavated clay will be stockpiles and then used as trench backfill. The clay is to be compacted into the trench in uniform depth layers at a total density of 1.8 t/m3 and a moisture content of 14%. After all the excavated clay is compacted into the trench what depth of the trench will still remain to be filled?
5- A rectangular excavation measuring 20 m by 40 m in plan is taken down to a depth of 3 m. The sides of the excavation have vertical faces. A cylindrical sample of the soil, with a diameter of 50 mm and height of 100 mm, is taken prior to excavation. The soil sample has an original mass of 363 g and mass of sample after oven drying is 310 g. The excavated soil is to be used to backfill a trench 1 m wide by 2 m deep. The soil is to be compacted into the trench at a moisture content of 24% such that its dry density is 1.6 t/m3 . a) What would be the shrinkage limit of the soil in the tube sample? b) What length of trench could be backfilled by the excavated soil? c) What would be the degree of saturation of the compacted soil in the trench?
Explanation / Answer
1) SW would be preferred over SP as a fill material because being well graded sand, SW will have particles of all sizes and thus would have more dense arrangement possible on compaction.
2) soil A is well graded gravel and soil B is highly plastic clay thus soil B would be more plastic whereas soil A would be more permeable.
3) e= 0.6 and S= 0.75 thus volume of voids = Vv = 0.6* Vs ( volume of solids)
S= Vw/Vv thus volume of water = 0.75* volume of voids
Total unit weight of soil = weight of water + weight of solids
Assuming Gs =2.7 we will have eS= wGs thus w= 0.167
Thus unit weight = Gs Yw (1+ w)/(1+e) = 19.69 KN/m3
Shrinkage limit of soil = (w-(V wet soil - V dry soil)*Yw/weight of dry soil)*100 = (w-(volume of water*Yw/ weight of dry soil)*100= (16.7-(1/Gs)) % = 16.329 %
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